You need $\hat\theta$ to be unbiased for $\theta$, which gives the constraint $\sum\limits_{i=1}^k a_i=1$ assuming $W_i$'s are unbiased for $\theta$. Subject to this restriction, you have to minimize the variance of $\hat\theta$.
Now,
\begin{align}
\operatorname{Var}_{\theta}(\hat\theta)&=\sum_{i=1}^k a_i^2\operatorname{Var}_{\theta}(W_i)+\sum_{i\ne j}\operatorname{Cov}_{\theta}(W_i,W_j)
\\&=\sum_{i=1}^k a_i^2\sigma_i^2
\end{align}
So you have the optimization problem (with respect to $a_1,\ldots,a_k$):
$$\text{Minimize} \quad \sum_{i=1}^k a_i^2\sigma_i^2\quad\text{subject to }\sum_{i=1}^k a_i=1$$
To solve this using Lagrange multipliers, you might take a look at the worked out examples here and here. This is a straightforward application of the method.
But you can alternatively use Cauchy-Schwarz inequality to directly say that
$$\left(\sum_{i=1}^k a_i^2\sigma_i^2\right) \left(\sum_{i=1}^k \frac{1}{\sigma_i^2}\right)\ge \left(\sum_{i=1}^k a_i\right)^2$$
Here equality holds if and only if $a_i\sigma_i=\frac{c}{\sigma_i}\implies a_i=\frac{c}{\sigma_i^2}$ for all $i$ and for some constant $c(\ne 0)$.
Since $\sum\limits_{i=1}^k a_i=1$, you have $c=\left(\sum\limits_{i=1}^k \frac{1}{\sigma_i^2}\right)^{-1}$. So equality is attained exactly at $\hat a_i=\frac{1/\sigma_i^2}{\sum\limits_{i=1}^k (1/\sigma_i^2)}$.
The estimator $W^*=\sum\limits_{i=1}^k \hat a_i W_i$ is what is called the best linear unbiased estimator (BLUE) of $\theta$.
