1. **Problem statement:** Find the coefficient of $x^4$ in the expansion of $$\left(\frac{x}{2} - \frac{3}{x^2}\right)^{10}.$$\n\n2. **Formula used:** We use the Binomial Theorem for expansion: $$\left(a+b\right)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k.$$\nHere, $a = \frac{x}{2}$ and $b = -\frac{3}{x^2}$, and $n=10$.\n\n3. **Find the general term:** The $k$-th term is $$T_{k+1} = \binom{10}{k} \left(\frac{x}{2}\right)^{10-k} \left(-\frac{3}{x^2}\right)^k = \binom{10}{k} \frac{x^{10-k}}{2^{10-k}} (-1)^k \frac{3^k}{x^{2k}} = \binom{10}{k} (-1)^k \frac{3^k}{2^{10-k}} x^{10-k-2k} = \binom{10}{k} (-1)^k \frac{3^k}{2^{10-k}} x^{10-3k}.$$\n\n4. **Find $k$ such that the power of $x$ is 4:**\nWe want $$10 - 3k = 4 \implies 3k = 6 \implies k = 2.$$\n\n5. **Calculate the coefficient for $k=2$:**\n$$\binom{10}{2} (-1)^2 \frac{3^2}{2^{8}} = 45 \times 1 \times \frac{9}{256} = \frac{405}{256}.$$\n\n6. **Final answer:** The coefficient of $x^4$ in the expansion is $$\boxed{\frac{405}{256}}.$$