Let 100 ml of one solution be mixed with 100 ml of other solution.

Mass of 100 ml of 30% $H_2SO_4 = 100 \times 1.218 = 121.8 g$

Mass of $H_2SO_4$ in 121.8 g of 30% $H_2SO_4 = \large\frac{30}{100}$$\times 121.8=36.54g$

Mass of water present in the solution = (121.8 – 36.54) = 85.26 g

Mass 100 ml of 70% $H_2SO_4 = 100 \times 1.61 = 161 g$

Mass of $H_2SO_4$ in 161.0 g of 70% $H_2SO_4 =\large\frac{70}{100}$$\times 161.0=112.7g$

Mass of water present in the solution = 161 – 112.7 = 48.3 g

Total mass of water in solution = (85.26 + 48.3) = 133.56 g =$\large\frac{133.56}{1000}$kg

Molality =Total no.of moles of $H_2SO_4$/Total mass of water in kg=$\large\frac{149.24}{98}\times \frac{1000}{133.56}$$=11.4moles/kg$

Hence (A) is the correct answer.