How to find the average half life of radioactive nuclide which undergoes two different decays?
But I am getting the correct answer only if I take $\lambda_2$ as $2$ times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?


Yes.



The way I tend to approach half-life problems is to recast them as the relevant kinetic differential equations.  Assuming the first-order kinetics as stated in the problem, the contribution of reaction 1 is
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_1 = -\lambda_1[A],
$$
and that of reaction 2 is&dagger;
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_2 = -2\lambda_2[A].
$$

Implicit in the above is the assumption that the first-order decay constants apply to the stoichiometry of the reactions as-written. As you correctly note, that extra factor of $2$ in the differential equation for reaction 2 is because two $A$ nuclei are involved per "unit" of reaction progression.

The total rate of loss of $A$ is the sum of the above:
$$
\frac{\mathrm{d}[A]}{\mathrm{d}t} 
= -\lambda_1[A] -2\lambda_2[A] 
= -\left(\lambda_1 +  2\lambda_2\right)[A].
$$

This is a straightforward first-order ODE, readily solved for the overall half-life by standard methods.



&dagger; While it seems quite unusual to me to have a second-order reaction ($2A\longrightarrow B$) with a first-order rate constant (units of $\mathrm{s}^{-1}$), for the sake of the problem I'm content to roll with it. I'm sure they cast the problem this way so that the calculus was straightforward.

As well, simultaneously having $A \longrightarrow B$ and $2A \longrightarrow B$ processes would seem to violate all kinds of conservation laws--the problem makes no physical sense as a result. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. $2A \longrightarrow C$ would have been much less preposterous, even if implausible in the context of low-energy radioactive decay.