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 \( \beta_1=0.359 \).