Initially for Argon $P\times V = \large\frac{4}{40}\times R\times T$

(molecular weight of Ar = 40)

On heating,

$P\times V = \large\frac{3.2}{40} \times R\times (T+50)$

$\therefore 4T = 3.2T + 160$

$T = \large\frac{160}{0.8}$

$\therefore\;\;\;\;=200K$

Hence answer is (c)