Writing the Cost Function

We define the MSE without the $1/n$ factor and have then, using that

$$\boldsymbol{X}\boldsymbol{\beta}=\begin{bmatrix} 2\beta_0 \\ \beta_1 \\0 \end{bmatrix},$$ $$C(\boldsymbol{\beta})=(4-2\beta_0)^2+(2-\beta_1)^2+\lambda(\beta_0^2+\beta_1^2),$$

and taking the derivative with respect to $\beta_0$ we get

$$\beta_0=\frac{8}{4+\lambda},$$

and for $\beta_1$ we obtain

$$\beta_1=\frac{2}{1+\lambda},$$

Using the constraint for $\beta_0^2+\beta_1^2=1$ we can constrain $\lambda$ by solving

$$\left(\frac{8}{4+\lambda}\right)^2+\left(\frac{2}{1+\lambda}\right)^2=1,$$

which gives $\lambda=4.571$ and $\beta_0=0.933$ and $\beta_1=0.359$.

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