Compute $\lim_{t\rightarrow0}\int_B \frac{l(t,x,y)}{t^2}d(x,y)$.

Update

The limit in question is $4\pi R$, where $R$ denotes the radius of the large disk. Here is a quick solution:

For fixed $t\ll R$ all the action is in the narrow zone $A:=\{(x,y)\>|\>R-t\leq\sqrt{x^2+y^2}\leq R\}$ immediately adjacent to the boundary circle $\partial B$. Furthermore the tiny circles of radius $t$ cannot "feel" the curvature of the huge circle $\partial B$. We can therefore model $\partial B$ by the $y$-axis modulo $2\pi R$ and let the strip $A'\colon \>0\leq x\leq t$ model the annulus $A$. One then has $$\ell(t,x,y)=2 t\,\phi= 2 t\>\arccos{x\over t}\qquad(0\leq x\leq t)$$ and therefore $$\int_{A'}\ell(t,x,y)\>{\rm d}(x,y)=2\pi R\cdot 2 t\int_0^t\arccos{x\over t}\>dx=4\pi R t^2\int_0^1\arccos\tau\>d\tau=4\pi R t^2\ .$$ This proves the claim. The argument shows that the limit in question is $2L$ for any smooth curve of length $L$.

I'm appending the original solution which works with the "real" $B$:

Assume that the radius of the big disk $B$ is $1$, and fix a positive $t\ll1$. For a point $(x,y)\in B$ with $\sqrt{x^2+y^2}=:r$ the length $\ell(t,x,y)$ is given by $\ell(t,x,y)=2t\phi$, whereby the involved data are related by $$r^2+t^2+2rt\cos\phi=1\ .$$ We consider $\phi$ as primary variable. Then $$r(\phi)=\sqrt{1-t^2\sin^2\phi}-t\cos\phi\ .$$ The relevant intervals are $0\leq\phi\leq{\pi\over2}+\arcsin{t\over2}$ and correspondingly $1-t\leq r(\phi)\leq1$ (draw a figure!). It follows that $$\int_B\ell(t,x,y)\>{\rm d}(x,y)=2t\int_0^{\pi/2+\arcsin(t/2)}\phi\>2\pi r(\phi)r'(\phi)\>d\phi\ .$$ Expanding with respect to $t$ Mathematica computes $$r(\phi)=1+?t,\qquad r'(\phi)=t(\sin\phi+?t)\ .$$ It follows that $$\int_B\ell(t,x,y)\>{\rm d}(x,y)=4\pi t^2\left(\int_0^{\pi/2}\phi\sin\phi\>d\phi+?t\right)=4\pi t^2(1+?t)\ ,$$ hence the claim.

(I write $?t^k$ for the remainder terms Mathematica abbreviates with $O[t]^k$.)