Need sql query for matching with three values

You can use GROUP BY with JOIN :

select t.*
from table t inner join
     ( select cid 
       from table
       where NetworkID in (1,2,3)
       group by cid
       having count(distinct NetworkID) = 3
     ) tt 
     on tt.cid  = t.cid; 

Try this:

select * from my_table t
where exists(select 1 from my_table
             where CID = t.CID and NetworkID in (1,2,3)
             group by CID
             having count(*) = 3)

Try this:

select * from <<tablename>> where cid in(select cid from <<tablename>> group by cid having count(*)=3).

Here the subquery will return you all thouse cid which have 3 rows in your table.

Or if you have more network ids then use of INTERSECT operator can be helpful:

select * from <<tablename>> where cid in (
   select cid from <<tablename>> where NetworkID=1
   INTERSECT 
   select cid from <<tablename>> where NetworkID=2
   INTERSECT
   select cid from <<tablename>> where NetworkID=3
);

INTERSECT operator basically returns all the rows common in the queries. Thus, your data unpredicatbility can be handled in this way

Try xml path.

SELECT *
FROM Table_Name B
WHERE (SELECT [text()] = A.Network FROM Table_Name A WHERE A.CID = B.CID 
ORDER BY CID, CAAccount FOR XML PATH('')) = 123

CTE Demo:

; WITH CTE(CAAccount, CID, Network) AS
(
SELECT 1 , 1, 1 UNION ALL
SELECT 2 , 1, 2 UNION ALL
SELECT 3 , 2, 1 UNION ALL
SELECT 4 , 2, 2 UNION ALL
SELECT 5 , 2, 3 UNION ALL
SELECT 6 , 3, 1 UNION ALL
SELECT 7 , 3, 2 UNION ALL
SELECT 8 , 3, 3 UNION ALL
SELECT 9 , 4, 1 UNION ALL
SELECT 10, 4, 2
) SELECT *
FROM CTE B
WHERE (SELECT [text()] = A.Network FROM CTE A WHERE A.CID = B.CID ORDER BY CID, CAAccount FOR XML PATH('')) = 123

Output:

CAAccount   CID Network
3           2   1
4           2   2
5           2   3
6           3   1
7           3   2
8           3   3