Suppose x2 occurs in the (r + 1)th term of the expansion (1+x)m.
Now Tr+1 =

m

‌

Crxr
On comparing power of x in x2 and Tr+1, we get r = 2.
Thus coefficient of x2 =

m

‌

C2
⇒

m

‌

C2 = 6 (given) ⇒

m!

2!(m−2)!

= 6
⇒ m (m – 1) = 12 ⇒ m2 – m – 12 = 0 ⇒ (m + 3) (m – 4) = 0 ⇒ m = – 3, 4
⇒ m = 4 (Since m cannot be negative)