If we try to write the regular expression of the language accepted by above NFA,it looks something like :
$\epsilon + 01^{*} + (01)^{*}$
From the above expression we can say that language L(M) accepts all strings starting with zero or the null string.Or, equivalently in set-builder notation:
$L(M)=\left \{ w:w\epsilon (\left \{ 0,1 \right \}^{*}\wedge {w} \ starts\ with\ zero \vee it's\ a\ null\ string) \right \}$
$\sim L(M)=\left \{ w: w \in \left \{ 0,1 \right \}^{*}\wedge w\ starts\ with\ 1 \right \}$
$So,regex(\sim L(M))=1(0+1)^{*}$