Let current of 1 ampere is passed through the solution for 1 hours.
$\therefore$ charge passed=$1\times 60\times 60=3600$
Mole of $e^-$ passed =$\large\frac{3600\times I}{96500}$
At anode :$2ON^-\rightarrow \large\frac{1}{2}$$O_2+H_2O+2e^-$
At cathode :$2H^++2e^-\rightarrow H_2$
Moles of $O_2$ released at anode $=\large\frac{3600\times I}{96500}\times \frac{1}{4}$
Volume of $O_2$ releases =$\large\frac{3600\times I}{96500}\times \frac{1}{4}$$\times 22400$
$\Rightarrow 208.9I$
Moles of $H_2$ released at cathode $=\large\frac{3600\times I}{96500}\times \frac{1}{2}$
Volume of $H_2$ releases =$\large\frac{3600\times I}{96500}\times \frac{1}{2}$$\times 22400$
$\Rightarrow 417.8I$
Total volume of gas released at S.T.P =208.9I+417.8I=338ml
I=0.5amp
Hence (d) is the correct answer.