We require to use Grahams Law of Diffusion to find a relationship between the volumes of the two gases and their molar masses.

When two gases are kept under the same conditions for pressure and temperature, the ratio that exists between their number of moles is equivalent to the ratio that exists between their volumes.

You can prove this by using the ideal gas equation equation - 

PV = nRT

where,

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821\frac{\text{atm}\cdot \text{L}}{\text{mol}\cdot \text{K}}$
$T$ - the absolute temperature of the gas

So, we can say, 

P x V_A = n_ART, for gas A

P x V_B = n_BRT, for gas B

Dividing these two equations we get,

V_A/V_B = n_A/n_B

This means that the diffusion of a volume of gas is equivalent to the diffusion of number of moles of gas.

Now, according to Grahams Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its density.

Using the ideal gas law, the we can write the equation for the density of a gas as a function of its molar mass.

n = m x M

Here $m$ is mass and M is molar mass. You will thus have

PV= mMRT

Rearranging it, we get P x M= mRT/V

P x M/RT = m x RT/V

P x M/RT = m/v 

So, the rate of diffusion can be expressed in terms of the volumes of the two gases and of their molar masses

Rate of diffusion of A = 1/√P x M_A/RT and Rate of diffusion of B = 1/√P x M_B/RT

Now, Rate of diffusion of A/Rate of diffusion of B = √d_B/√dA  

(At constant temp. and pressure)

So, Rate of diffusion of A/Rate of diffusion of B = √M_B x P₁/ √M_A x P₂

Therefore, (V_A x t_B/V_B x t_A) = √dB/√dA = √M₂/√M_A

Since we know that,

Rate of diffusion of A/Rate of diffusion of B = √M₂/√M_{A =} (180 x 10/60 x 15) 

√M₂/√M_A = 2 

Now, √M₂/√16 = 2

So, √M₂/4 = 2

Then, √M₂ = 8

And, M = √8 

= 2.83 U