Solution: We recall that the Picard's iteration for the initial value problem \[ \frac{\mathrm{d}y}{\mathrm{d}x} = f(x, y), \quad y(x_0) = y_0 \] is given by the iterative formula \[ y_{n+1}(x) = y_0 + \int_{x_0}^{x} f(t, y_n(t)) \, \mathrm{d}t. \] For the given problem, we have \[ \frac{\mathrm{d}y}{\mathrm{d}x} = 2y - 2x^2 - 3, \quad y(0) = 2. \] Thus, the iterative formula becomes \[ y_{n+1}(x) = 2 + \int_{0}^{x} \left( 2y_n(t) - 2t^2 - 3 \right) \, \mathrm{d}t. \]

Starting with the initial approximation \( y_0(x) = 2 \), we compute the next approximations as follows:

First Approximation: $y_0(x) = 2$. \begin{align*} y_{1}(x) & = 2 + \int_{0}^{x} \left( 2y_0(t) - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 2 \cdot 2 - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 4 - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 1 - 2t^2 \right) \, \mathrm{d}t\\ & = 2 + \left[ t - \frac{2t^3}{3} \right]_{0}^{x} \\ & = 2 + \left( x - \frac{2x^3}{3} \right) \\ & = 2 + x - \frac{2x^3}{3}. \end{align*}

Second Approximation: $y_1(x) = 2 + x - \frac{2x^3}{3}.$ \begin{align*} y_2(x) & = 2 + \int_{0}^{x} \left( 2 \left( 2 + t - \frac{2t^3}{3} \right) - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 4 + 2t - \frac{4t^3}{3} - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 1 + 2t - 2t^2 - \frac{4t^3}{3} \right) \, \mathrm{d}t\\ & = 2 + \left[ t + t^2 - \frac{2t^3}{3} - \frac{t^4}{3} \right]_{0}^{x}\\ & = 2 + x + x^2 - \frac{2x^3}{3} - \frac{x^4}{3}. \end{align*}

Third Approximation: $y_2(x) = 2 + x + x^2 - \frac{2x^3}{3} - \frac{x^4}{3}.$ \begin{align*} y_3(x) & = 2 + \int_{0}^{x} \left( 2 \left( 2 + t + t^2 - \frac{2t^3}{3} - \frac{t^4}{3} \right) - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 4 + 2t + 2t^2 - \frac{4t^3}{3} - \frac{2t^4}{3} - 2t^2 - 3 \right) \, \mathrm{d}t \\ & = 2 + \int_{0}^{x} \left( 1 + 2t - \frac{4t^3}{3} - \frac{2t^4}{3} \right) \, \mathrm{d}t \\ & = 2 + \left[ t + t^2 - \frac{t^4}{3} - \frac{2t^5}{15} \right]_{0}^{x} \\ & = 2 + x + x^2 - \frac{x^4}{3} - \frac{2x^5}{15}. \end{align*} Thus, the third approximation is \[ \textcolor{blue}{\boxed{ y_3(x) = 2 + x + x^2 - \frac{x^4}{3} - \frac{2x^5}{15} }} \]