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 .