Let us use Grahams law to answer this question-

Mol wt of methane = 16 g/mol

Let the mol wt of the gas which effuses two times as long to effuse be x

Then, according to Graham law, 

r₁ over r₂ = √MM₂ over √MM₁ , where r₁ is the rate of effusion of methane and r₂ is that for the other gas. So if r₁ = 1, then r₂ = 2 x r₁.

So, 1/2 = √x/√16

So, 1/2 = √x/4

√x = 4/2 = 2

x = √2

= 1.414, which is approximately the atomic mass of Hydrogen.