Step 1: List the probabilities.
We have $P(0)=\tfrac{5}{16}$, $P(1)=\tfrac{5}{16}$, $P(2)=\tfrac{k\cdot 2}{48}=\tfrac{k}{24}$, and $P(3)=\tfrac{1}{4}$.
Step 2: Use the total rule.
All probabilities of a p.m.f. must add to $1$. So $\tfrac{5}{16}+\tfrac{5}{16}+\tfrac{k}{24}+\tfrac{1}{4}=1$.
Step 3: Solve for k.
The fixed parts add to $\tfrac{5}{16}+\tfrac{5}{16}+\tfrac{4}{16}=\tfrac{14}{16}=\tfrac{7}{8}$. So $\tfrac{k}{24}=1-\tfrac{7}{8}=\tfrac{1}{8}$, giving $k=3$.
Step 4: Find each probability as sixteenths.
Now $P(2)=\tfrac{3\cdot 2}{48}=\tfrac{1}{8}=\tfrac{2}{16}$. So in sixteenths: $P(0)=\tfrac{5}{16}$, $P(1)=\tfrac{5}{16}$, $P(2)=\tfrac{2}{16}$, $P(3)=\tfrac{4}{16}$.
Step 5: Apply the mean formula.
$E(X)=\sum x\,P(x)=0\cdot\tfrac{5}{16}+1\cdot\tfrac{5}{16}+2\cdot\tfrac{2}{16}+3\cdot\tfrac{4}{16}$.
Step 6: Add it up.
This is $\tfrac{5+4+12}{16}=\tfrac{21}{16}=1.3125$, which is option (B).
\[ \boxed{\,E(X)=1.3125\,} \]