Step 1:
Let $V$ml of each are mixed.
$H_2SO_4$ is 30% by weight
Wt of $H_2SO_4=30g$
Wt of solution =100g
Volume of solution=$\large\frac{100}{1.218}$$ml$
(i.e) $\large\frac{100}{1.218}$$ml$ contains 30g $H_2SO_4$
$\therefore$ Vml contains $\large\frac{30\times V\times 1.218}{100}$ of $H_2SO_4$
Step 2:
$H_2SO_4$ is 70% by wt.
$\therefore$ Wt of $H_2SO_4$=70g
Wt of solution =100g
Volume of solution =$\large\frac{100}{1.610}$ml
(i.e) $\large\frac{100}{1.610}$ml contains 70g $H_2SO_4$
$V$ml contains $\large\frac{70\times V\times 1.610}{100}$g $H_2SO_4$
On mixing these two,total wt of $H_2SO_4=\bigg[\large\frac{30\times 1.218}{100}+\frac{70\times 1.610}{100}\bigg]$$Vg$
$\Rightarrow 1.4924Vg$
Step 3:
Total volume of solution =2Vml
Molarity of solution =$\large\frac{1.4924V}{98\times \Large\frac{2V}{1000}}$
$\Rightarrow 7.61$
Now wt of total solution $=2V\times 1.425g$
$\Rightarrow 2.85Vg$
Wt of water =(2.85V-1.4924V)g
$\Rightarrow 1.3576Vg$
Molality of solution =$\large\frac{1.4924V}{98\times \Large\frac{1.3576V}{1000}}$
$\Rightarrow 11.22$
Hence (b) is the correct option.