jmc

Let $g(x)=f(x)-x.$ Then $g(x)$ is a cubic polynomial, and $g(0)=g(1)=g(2)=0$ and $g(3)=1,$ so $$g(x)=kx(x-1)(x-2)$$for some constant $k.$ Setting $x=3,$ we get $$g(3)=k(3)(2)(1),$$so $6k=1.$ Hence, $k=\frac{1}{6},$ so $$g(x)=\frac{x(x-1)(x-2)}{6},$$and $f(x)=\frac{x(x-1)(x-2)}{6}+x.$ In particular, $f(5)=\frac{(5)(4)(3)}{6}+5=\boxed{15}.$