Because it doesn't seem to be possible to solve for $w(t)$, what you could do to obtain the plot of $w(t)$ is by numerically solving for $w$ for different values of $t$ with FindRoot. The results can then be plotted with ListLinePlot.

numberOfPoints = 300;
wStart = 1;
ts = Rest @ Subdivide[0, 3, numberOfPoints];
ws = Last @* Last /@ Table[
  FindRoot[Evaluate[eqn /. t -> n], {w, wStart}],
  {n, ts}
];
ListLinePlot[Transpose @ {ws, ts}]

(Note that Rest is to exclude the case in which $t = 0$.)

Update

At first I ignored the FindRoot::lstol messages, which I shouldn't have. One way of checking if each pair of numerically-obtained $(w, t)$ actually satisfies eqn is by looking at the difference between its left and right sides. The difference should be close to zero.

Trying this:

numberOfPoints = 5;
wStart = 1;
ts = Rest @ Subdivide[0, 3, numberOfPoints];
ws = Last @* Last /@ Table[
  FindRoot[Evaluate[eqn /. t -> n], {w, wStart}],
  {n, ts}
];
points = Transpose @ {ws, ts};
diff = Subtract @@ List @@ eqn;
diff /. {w -> First @ #, t -> Last @ #} & /@ points

we get:

{-4.72339, -14.1116, -26.7086, -41.0978, -55.888}

As they're not relatively close to zero, the numerical solutions are bad. One way of addressing this is by giving FindRoot a better wStart.

To find it, we could explore the plot of the difference at some values of t to visually identify w at which diff is zero. It appears that

Plot[Evaluate[Table[diff /. t -> n, {n, ts}]], {w, -3, 10}, PlotRange -> {-600, 50}]

which signifies that wStart should be somewhere near -2 for $t \in (0, 3]$. Let's try wStart = -3:

numberOfPoints = 5;
wStart = -3;
ts = Rest @ Subdivide[0, 3, numberOfPoints];
ws = Last @* Last /@ Table[
  FindRoot[Evaluate[eqn /. t -> n], {w, wStart}],
  {n, ts}
];
points = Transpose @ {ws, ts};
diff = Subtract @@ List @@ eqn;
diff /. {w -> First @ #, t -> Last @ #} & /@ points

from which we get

{2.226*10^-13, 4.9738*10^-14, 4.36984*10^-13, 7.31859*10^-13, -4.54747*10^-13}

which are all close to zero, so this wStart is good. Now, back to the code at the beginning, with the better wStart (and note that because t shouldn't be too close to zero, I let ts start at 11/100 rather than 0):

numberOfPoints = 300;
wStart = -3;
ts = Subdivide[11/100, 3, numberOfPoints];
ws = Last @* Last /@ Table[
  FindRoot[Evaluate[eqn /. t -> n], {w, wStart}],
  {n, ts}
];
ListLinePlot[Transpose @ {ws, ts}]

No more FindRoot::lstol messages.

A way:

eqn = t^3 + (4 - 8/Sqrt[\[Pi]]) w^2 + (3 w^4)/
8 + (2 E^(w^2/t) \[Pi] w^2 (-2 t + w^2) Erfc[
    w/(Sqrt[2] Sqrt[t])]^2)/
t - (1/8 t (96 + 56 t + 
    11 w^2) + (E^(w^2/(2 t)) w (-64 w^2 + 
      Sqrt[\[Pi]] (t (-160 + 7 t) - 8 (-8 + t) w^2 + 3 w^4)) Erfc[
     w/(Sqrt[2] Sqrt[t])])/(8 Sqrt[2] Sqrt[t])) // Simplify

ContourPlot[eqn == 0, {t, 1/10, 3}, {w, -3, 3}, FrameLabel -> Automatic]

In near point t=0

ContourPlot[eqn == 0, {t, 1/10000, 3}, {w, -3, 3}, FrameLabel -> Automatic,
PlotPoints -> 100, MaxRecursion -> 4, WorkingPrecision -> 20]

Real part of eqn:

 Show[{ContourPlot[(eqn // Re) == 0, {t, -2, -1/100}, {w, -3, 3}, 
 FrameLabel -> Automatic, WorkingPrecision -> 20], 
 ContourPlot[(eqn // Re) == 0, {t, 1/100, 3}, {w, -3, 3}, 
 FrameLabel -> Automatic, WorkingPrecision -> 20]}, 
 PlotRange -> {{-3, 3}, {-3, 3}}]