Given,

$A=\{x\in ℝ:\left[x+3\right]+\left[x+4\right]\le 3\} \& B=\left\{x\in ℝ:3^x{\left(\sum _{r=1}^{\infty }\frac{3}{{10}^r}\right)}^{x-3}<3^{-3x}\right\},$

Now solving $A$ we get,

$\left[x+3\right]+\left[x+4\right]\le 3$

$\Rightarrow \left[x\right]+3+\left[x\right]+4\le 3$

$\Rightarrow 2\left[x\right]\le -4$

$\Rightarrow \left[x\right]\le -2$

$\Rightarrow x<-1\Rightarrow A=\left(-\infty ,-1\right)$

Now solving $B$ we get,

$3^x{\left(\sum _{r=1}^{\infty }\frac{3}{{10}^r}\right)}^{x-3}<3^{-3x}$

Now using infinite $\mathrm{G}.\mathrm{P}$ formula $S_{\infty }=\frac{a}{1-r}$ we get,

$\Rightarrow 3^x{\left(\frac{3·\frac{1}{10}}{1-\frac{1}{10}}\right)}^{x-3}<3^{-3x}$

$\Rightarrow 3^x{\left(\frac{1}{3}\right)}^{x-3}<3^{-3x}$

$\Rightarrow 3^{x-x+3+3x}<1$

$\Rightarrow 3^{3\left(x+1\right)}<1$

$\Rightarrow 3\left(x+1\right)<0$

$\Rightarrow x<-1\Rightarrow B=\left(-\infty ,-1\right)$

$\Rightarrow A=B$