Closure of regular languages under deleting a 1 from each even run of 1s
Below you can find solutions for the similar problem in which we want to delete a single 1 from each odd length run of 1s. I'll let you adapt them to your question.

First, let me explain how to solve it using closure properties. We start with your language $R$. We then apply a regular substitution that optionally "marks" every 1, i.e. changes it into $\mathbf 1$. We intersect the resulting language with $0^*((11)^*(\mathbf{1}+\epsilon)0^+)^*((11)^*(\mathbf1+\epsilon)+\epsilon)$ – this has the effect of marking the final 1 out of each odd run. Finally, we apply a homomorphism that deletes all marked 1s.

Second, let me explain how to explicitly construct an NFA for the new language given an NFA for $R$. Let the NFA for $R$ be $\langle Q, \{0,1\}, \delta, q_0, F \rangle$. The new states will be $(Q \times \{0,1\}) \cup \{q_f\}$, and the new initial state will be $\langle q_0, 0 \rangle$; the second entry is the parity of the current run of 1s. The new accepting states are $(F \times \{0\}) \cup \{q_f\}$ (note that in the new language, all runs of 1s must have even length). It remains to define the new transition function:


If there is a transition $q_1 \stackrel 0 \to q_2$, then we add a transition $\langle q_1, 0 \rangle \stackrel 0 \to \langle q_2, 0 \rangle$; that is, we allow the transition only if the parity of the current run of 1s is even.
If there is a transition $q_1 \stackrel 1 \to q_2$, then we add the following transitions:


$\langle q_1, 0 \rangle \stackrel 1 \to \langle q_2, 1 \rangle$ and $\langle q_1, 1 \rangle \stackrel 1 \to \langle q_2, 0 \rangle$; these just maintain the parity of the length of the run.
If $q_2 \in F$: $\langle q_1, 0 \rangle \stackrel \epsilon \to q_f$. This handles odd runs of 1s at the very end of input.

If there are transitions $q_1 \stackrel 1 \to q_2 \stackrel 0 \to q_3$, then we add the transition $\langle q_1, 0 \rangle \stackrel 0 \to \langle q_3, 0 \rangle$. This corresponds to deleting a 1 from the end of an odd run.