Given, $Na_2S_2O_3$ has molarity = $3mol\;litre^{-1}$

Moles of $Na_2S_2O_3$ = 3

$\therefore$ Weight of $Na_2S_2O_3 = 3\times158 = 474g$

and, Volume of the solution = 1 litre = 1000mL

$\therefore$ Weight of the solution = $1000\times1.25$ = 1250 g

$\therefore$ weight of water = 1250 - 474 = 776 g

$\%$ by weight of $Na_2S_2O_3$ =$\large\frac{Wt.\;of\; Na_2S_2O_3}{Wt.\; of \;solution}\times100$ $= \large\frac{474}{1250}\times100 = 37.92$

Hence answer is (c)