Let the first ball's initial speed be $u_1$.
It is thrown vertically upward, so its total time of flight is $T_1 = \frac{2u_1}{g}.$

Let the second ball's initial speed be $u_2$.
It is thrown at an angle $\theta$ with the vertical, so its vertical component is $u_{2y} = u_2 \cos\theta.$ Hence its time of flight is $T_2 = \frac{2u_2\cos\theta}{g}.$

Given both balls stay in air for the same time: $T_1 = T_2$ $\frac{2u_1}{g} = \frac{2u_2\cos\theta}{g}$ $u_1 = u_2\cos\theta.$

Maximum height of the first ball: $H_1 = \frac{u_1^2}{2g}.$

Maximum height of the second ball:
Only the vertical component matters for maximum height: $H_2 = \frac{(u_2\cos\theta)^2}{2g}.$

Using $u_1 = u_2\cos\theta$ from step 3, $H_2 = \frac{u_1^2}{2g} = H_1.$

Therefore, $\frac{H_1}{H_2} = 1.$

The question says the ratio of heights attained by the two balls respectively is $\frac{1}{x}.$ So, $\frac{1}{x} = 1 \implies x = 1.$

Note: The masses $m$ and $2m$ do not affect projectile motion here, so they are irrelevant.