4. $\frac{1}{2}$.

Since $f(x)=\left|x^{2}-a\right|$ is an even function,
$\therefore M(a)$ is the maximum value of $f(x)$ in the interval $0 \leqslant x \leqslant 1$.
When $a \leqslant 0$, $f(x)=x^{2}-a$, it is easy to know that $M(a)=1-a$.
When $a>0$, from the graph we can see that.
If $\sqrt{2 a} \geqslant 1$, then $M(a)$ $=a$;

If $\sqrt{2 a} \leqslant 1$, then $M(a)$ $=f(1)=1-a$.
$$\therefore M(a)=\left\{\begin{array}{r} 1-a,\left(a \leqslant \frac{1}{2}\right) \\ a .\left(a \geqslant \frac{1}{2}\right) \end{array}\right.$$

Since $a \leqslant \frac{1}{2}$ when $M(a) \backslash$, and $a \geqslant \frac{1}{2}$ when $M(a) \lambda$. $\therefore a=\frac{1}{2}$ when, $\dot{M}(a)$ has the minimum value of $\frac{1}{2}$.