1. For an ideal gas, the diffusion coefficient is proportional to mean free path and mean speed:

$D \propto \lambda \, \bar{v}$

2. Mean free path of a gas is given by

$\lambda \propto \frac{T}{P}$

because

$\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}$

3. Mean speed is proportional to the square root of temperature:

$\bar{v} \propto \sqrt{T}$

4. Therefore,

$D \propto \lambda \, \bar{v} \propto \frac{T}{P}\cdot \sqrt{T} = \frac{T^{3/2}}{P}$

5. Now temperature is increased $4$ times and pressure is increased $2$ times:

$\frac{D_2}{D_1} = \frac{(4T)^{3/2}/(2P)}{T^{3/2}/P}$

6. Simplify:

$\frac{D_2}{D_1} = \frac{4^{3/2}}{2} = \frac{8}{2} = 4$

So,

$x = 4$