Your strategy is correct: the formula you cite, $$\require{enclose} \mathring e_x = \mathring e_{x : \enclose{actuarial}{n}} + {}_n p_x \; \mathring e_{x + n} \tag{1}$$ with the choice $n = 3$, yields $$\mathring e_{x : \enclose{actuarial}{3}} = \mathring e_x - {}_3 p_x \; \mathring e_{x+3}. \tag{2}$$ Since the only quantity that we are not given is ${}_3 p_x$, we have to compute it. Recall that $${}_n p_x = 1 - {}_n q_x, \tag{3}$$ that is to say, the probability that $(x)$ survives $n$ years is equal to the complement of the probability that $(x)$ dies within $n$ years. Since $q_x$ is the probability that $(x)$ dies within the next year, and ${}_2 q_{x+1}$ is the probability that $(x+1)$ dies within the next two years, the probability that $(x)$ dies within the next $3$ years is $$\begin{align}
{}_3 q_x &= q_x + (p_x)({}_2 q_{x+1}) \\
&= q_x + (1 - q_x)({}_2 q_{x+1}). \tag{4} \end{align}$$ This is because either $(x)$ dies in the next year, or $(x)$ survives the first year to reach $(x+1)$ but dies within the next two years. Putting $(3)$ together with $(4)$ then gives us
$$\begin{align}
{}_3 p_x &= 1 - q_x - (1 - q_x)({}_2 q_{x+1}) \\
&= (1 - q_x)(1 - {}_2 q_{x+1}). \tag{5}
\end{align}$$
We could also have obtained the identity $(5)$ by observing that $${}_3 p_x = (p_x)({}_2 p_{x+1}), \tag{6}$$ since $(x)$ survives $3$ years if they survive the first year to reach $(x+1)$, then survives an additional $2$ years. Then we apply $(3)$ to each factor on the right hand side, and we again get $(5)$.
All that remains is to substitute $(5)$ into $(2)$ and use the given information to compute the result, which I leave to you.
