Let $G$ be a central extension of the group $K$ by the group $H$. If we know that this extension is nonsplit, is it true that the order of $K$ must divide the Schur multiplier of the group $H$?

Just so we agree on the setup, you have an exact sequence $$ 1\rightarrow K\rightarrow G\rightarrow H\rightarrow 1, $$ where $K\leq Z(G)$, and you assume that the extension is nonsplit. You are asking whether it is true that $K$ divides the Schur multiplier of $H$. The answer is "no", you can make $K$ arbitrarily large. Indeed, having found one such extension, take $G\times A$ for an abelian group $A$. This is an extension of the form $$ 1\rightarrow K\times A\rightarrow G\times A\rightarrow H\rightarrow 1, $$ and clearly $K\times A\leq Z(G\times A)=Z(G)\times A$. Moreover, the extension still doesn't split, since the map $G\times A\rightarrow H$ factors through $G\rightarrow H$. That's precisely why in the theory of Schur multipliers you have to demand in addition that $K\leq G'$, and the above example shows that you cannot drop this condition. The above example had $(G\times A)' = G'$. So we no longer had $K\times A\leq G'$, even if this was true for $K$ itself. 

