Ans:$\dfrac{2}{15}$

We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(1,0),B=(1,1)$. We have $E=(\tfrac 12,0)$ and $F=(1,\tfrac 12)$. $\overline{BD}$ is governed by $y=x$. $\overline{BE}$ is governed by $y=2x-1$ and $\overline{AF}$ is governed by $y=-\frac 12x+1$. Now we can easily compute $P=(\tfrac 13,\tfrac 13),Q=(\tfrac 23,\tfrac 23),,R=(\tfrac 45,\tfrac 35)$. We finish with Shoelace.

We can also use Barycentric coordinates w.r.t. $\triangle DBC$. From above, we have that $DP=PQ=QB$. So we can immediately compute $D=(1,0,0),B=(0,1,0),C=(0,0,1),E=(\tfrac 12,0,\tfrac 12),F=(0,\tfrac 12,\tfrac12),P=(\tfrac 23,\tfrac 13,0),Q=(\tfrac 13,\tfrac23,0)$. The coordinates of $R$ are also easy enough to compute with equations of lines $\overline{QF}\cap\overline{BE}=R$. Then just compute $1-([DPE]+[BEC]+[BQR])$ with barycentric area formula.