Old circuit: series $R$-$L$

Given: $X_L = 3R$

For a series $R$-$L$ circuit, the impedance is $Z_1 = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (3R)^2} = \sqrt{10}\,R$

Hence the old power factor is $\cos\phi_1 = \frac{R}{Z_1} = \frac{R}{\sqrt{10}R} = \frac{1}{\sqrt{10}}$

New circuit: series $R$-$L$-$C$

A capacitor is added such that $X_C = 2R$

Net reactance becomes $X = X_L - X_C = 3R - 2R = R$

So new impedance is $Z_2 = \sqrt{R^2 + X^2} = \sqrt{R^2 + R^2} = \sqrt{2}\,R$

Therefore the new power factor is $\cos\phi_2 = \frac{R}{Z_2} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}}$

Ratio of new power factor to old power factor

$\cos\phi_2 : \cos\phi_1 = \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{10}}$

Multiply both terms by $\sqrt{20}$ (or simplify directly): $\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{10}} = \sqrt{5} : 1$

Comparing with the given ratio $\sqrt{5} : x$ we get $x = 1$

Final answer

$\boxed{1}$