1. The problem is to evaluate the magnitude of the complex number $\frac{5i}{3 - i}$.\n\n2. Recall that the magnitude (or modulus) of a complex number $z = a + bi$ is given by $|z| = \sqrt{a^2 + b^2}$.\n\n3. First, simplify the complex fraction $\frac{5i}{3 - i}$ by multiplying numerator and denominator by the conjugate of the denominator to remove the imaginary part from the denominator. The conjugate of $3 - i$ is $3 + i$.\n\n4. Multiply numerator and denominator:\n$$\frac{5i}{3 - i} \times \frac{3 + i}{3 + i} = \frac{5i(3 + i)}{(3 - i)(3 + i)}$$\n\n5. Calculate numerator:\n$$5i(3 + i) = 5i \times 3 + 5i \times i = 15i + 5i^2$$\nSince $i^2 = -1$, this becomes:\n$$15i + 5(-1) = 15i - 5$$\nSo numerator is $-5 + 15i$.\n\n6. Calculate denominator using difference of squares formula:\n$$(3 - i)(3 + i) = 3^2 - i^2 = 9 - (-1) = 9 + 1 = 10$$\n\n7. So the simplified form is:\n$$\frac{-5 + 15i}{10} = -\frac{5}{10} + \frac{15}{10}i = -\frac{1}{2} + \frac{3}{2}i$$\n\n8. Now find the magnitude of $-\frac{1}{2} + \frac{3}{2}i$:\n$$\left| -\frac{1}{2} + \frac{3}{2}i \right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{9}{4}} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}}$$\n\n9. Simplify the square root if desired:\n$$\sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}$$\n\n**Final answer:**\n$$\left| \frac{5i}{3 - i} \right| = \frac{\sqrt{10}}{2}$$