Step 1: Multiply both the numerator and the denominator of the given complex number by the conjugate of the denominator.

$$\dfrac{2}{i(3-i)} = \dfrac{2}{3i + 1} \times \dfrac{3i - 1}{3i - 1}$$

Step 2: Simplify the expression by applying the distributive property and the fact that $i^2 = -1$.

$$\dfrac{2(3i - 1)}{(3i + 1)(3i - 1)} = \dfrac{6i - 2}{9i^2 - 1} = \dfrac{6i - 2}{-9 - 1} = \dfrac{6i - 2}{-10}$$

Step 3: Separate the fraction into real and imaginary parts.

$$\dfrac{6i - 2}{-10} = -\dfrac{2}{10} + \dfrac{6}{10}i = -\dfrac{1}{5} + \dfrac{3}{5}i$$

Step 4: Express the result in the form $a + bi$.

$$\dfrac{2}{i(3-i)} = -\dfrac{1}{5} + \dfrac{3}{5}i = \boxed{\dfrac{1{-}3i}{5}}$$