The following assumes that $W$ and $a$ are positive integers, that one chooses uniformly at random $aW$ integers between $1$ and $2aW$, and that one asks for the probability of the event $A$ that the resulting set $S$ contains at least $W$ consecutive integers.
Let us compute the probability of the event $A_i$ that $S$ contains the $W$ consecutive integers starting from $i$. Then $A_i=\emptyset$ if $i\ge(2a-1)W+2$, otherwise $S$ is determined by its $(a-1)W$ elements not between $i$ and $i+W-1$. Hence $ P(A_i)=\frac{{(2a-1)W\choose (a-1)W}}{{2aW\choose aW}}=p(W). $ Similar formulas can be written for $P(A_i\cap A_j)$ for example, depending of $|i-j|$.
Since $A$ is the union of the events $A_i$, an exact formula for $P(A)$ is given by the exclusion-inclusion formula, and an easy upper bound is $ P(A)\le(2a-1)Wp(W). $ When $W\to+\infty$, Stirling formula yields $ p(W)=\sqrt{\frac{2a-1}{2a-2}}(C_a)^W(1+o(1)),\quad C_a=(2a-1)^{2a-1}2^{-2a}a^{-a}(a-1)^{1-a}. $ Roughly speaking, the error made in the uper bound is to count several times the segments of length greater than $W$ included in $S$. This remark and some heuristic arguments suggest that a discount of $\displaystyle\frac{a}{2a-1}$ should be applied to the upper bound, hence $ P(A)=a\sqrt{\frac{2a-1}{2a-2}}W(C_a)^W(1+o(1)). $