Star flips a quarter four times. Find the probability that the quarter lands heads exactly twice.

$\textbf{(A) }\dfrac18\hspace{14em}\textbf{(B) }\dfrac3{16}\hspace{14em}\textbf{(C) }\dfrac38$

$\textbf{(D) }\dfrac12$
1. **Determine the total number of possible outcomes:**
   - When flipping a quarter four times, each flip has 2 possible outcomes (heads or tails).
   - Therefore, the total number of possible outcomes is:
     \[
     2^4 = 16
     \]

2. **Determine the number of favorable outcomes:**
   - We need to find the number of ways to get exactly 2 heads in 4 flips.
   - This is a combination problem where we choose 2 positions out of 4 for the heads.
   - The number of ways to choose 2 positions out of 4 is given by the binomial coefficient:
     \[
     \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6
     \]

3. **Calculate the probability:**
   - The probability of getting exactly 2 heads in 4 flips is the number of favorable outcomes divided by the total number of possible outcomes:
     \[
     P(\text{2 heads}) = \frac{\binom{4}{2}}{2^4} = \frac{6}{16} = \frac{3}{8}
     \]

Conclusion:
\[
\boxed{\frac{3}{8}}
\]