We start by solving for $W$ in terms of $X_1$, $X_2$ and $X_3$:
$W = X_1 -aX_2 -bX_3$
Next, we compute the covariance between $W$ and $X_2$:
$Cov(W,X_2) = Cov(X_1 -aX_2 -bX_3,X_2) = Cov(X_1,X_2)-aV(X_2)-bCov(X_3,X_2)$
Here we used the fact that $Cov(\cdot,\cdot)$ is linear on each component and that $Cov(Z,Z)=Var(Z)$ for any random variable.
Next, we use the fact that $Cov(X_1,X_2) = 1.5$, $Var(X_2) = 1$, $Cov(X_3,X_2) = 0.5$ and set the last equation to zero so $W$ and $X_2$ are uncorrelated:
$Cov(W,X_2) = 1.5 - a-b\cdot0.5 = 0$
This gives us one equation in terms of $a$ and $b$. Next, we repeat the process with the covariance between $W$ and $X_3$:
$Cov(W,X_3) = Cov(X_1,X_3)-aCov(X_2,X_3)-bVar(X_3) = 1 -a\cdot 0.5-b = 0$
We then solve the system of two equations with two unknowns:
$1.5 - a-b\cdot0.5 = 0$
$1 -a\cdot 0.5-b = 0$
Solving for $a$ in the first equation, $a = 1.5 -b\cdot0.5$. Replacing in the second equation,
$1 -(1.5 -b\cdot0.5)\cdot 0.5-b = 0$,
which has as solution $b=\frac{1}{3}$, so that $a=\frac{4}{3}$.
The correct answer is then B. $\frac{4}{3}$
