Such ceiling

$\begin{aligned} A & = \underbrace{\left\lceil \sqrt 3\right\rceil + \left\lceil \sqrt 4\right\rceil}_{2 \sqrt 4} + {\color{#3D99F6} \underbrace{\left\lceil \sqrt 5\right\rceil + \cdots + \left\lceil \sqrt 9\right\rceil}_{(9-4)\sqrt 9} + \underbrace{\left\lceil \sqrt {10}\right\rceil + \cdots + \left\lceil \sqrt {16}\right\rceil}_{(16-9)\sqrt {16}} + \cdots + \underbrace{\left\lceil \sqrt {65} \right\rceil + \cdots + \left\lceil \sqrt {81} \right\rceil}_{(81-64)\sqrt {81}}} + \underbrace{\left\lceil \sqrt {82} \right\rceil + \left\lceil \sqrt {83} \right\rceil}_{2\sqrt {100}} \\ & = 2(2) + {\color{#3D99F6}\sum_{k=3}^9 \left(k^2-(k-1)^2\right)k} + 2(10) \\ & = 4 + \sum_{k=3}^9 \left(2k^2-k\right) + 20 \\ & = 4 + 2\times \frac {9(9+1)(2(9)+1)}6 - 2\times \frac {2(2+1)(2(2)+1)}6 - \frac {9(9+1)}2 + \frac {2(2+1)}2 + 20 \\ & = 4 + 570 - 10 - 45 + 3 + 20 \\ & = 542 \end{aligned}$

Therefore, $\dfrac 13 \left \lceil \sqrt A \right \rceil = \dfrac 13 \times 24 = \boxed{8}$ .

We know that for $(n-1)^2 < x \leq n^2$ , where $n$ is a positive integer,

$\lceil \sqrt{x} \rceil = n$

For example, $n=3, \;\; 2^2 < x \leq 3^3 \implies 4 < x \leq 9$

For any value of $x$ within this range, $\lceil \sqrt{x} \rceil = 3$

With this, we can easily calculate the value of the expression $A$

What we need to do is find out the number of terms between each perfect square

$\lceil \sqrt{3} \rceil =\lceil \sqrt{4} \rceil = 2$ , there are $2$ terms here

$\lceil \sqrt{5} \rceil =\lceil \sqrt{6} \rceil =\ldots = \lceil \sqrt{9} \rceil =3$ , there are $9-4 = 5$ terms here

$\lceil \sqrt{10} \rceil =\lceil \sqrt{11} \rceil =\ldots = \lceil \sqrt{16} \rceil = 4$ , there are $16-9 = 7$ terms here

$\lceil \sqrt{17} \rceil =\lceil \sqrt{18} \rceil = \ldots = \lceil \sqrt{25} \rceil =5$ , there are $25-16 = 9$ terms here

$\lceil \sqrt{26} \rceil =\lceil \sqrt{27} \rceil = \ldots = \lceil \sqrt{36} \rceil =6$ , there are $36-25=11$ terms here

$\lceil \sqrt{37} \rceil =\lceil \sqrt{38} \rceil = \ldots = \lceil \sqrt{49} \rceil =7$ , there are $49-36=13$ terms here

$\lceil \sqrt{50} \rceil =\lceil \sqrt{51} \rceil = \ldots = \lceil \sqrt{64} \rceil =8$ , there are $64-49=15$ terms here

$\lceil \sqrt{65} \rceil =\lceil \sqrt{66} \rceil = \ldots = \lceil \sqrt{81} \rceil =9$ , there are $81-64=17$ terms here

$\lceil \sqrt{82} \rceil =\lceil \sqrt{83} \rceil = 10$ , there are $2$ terms here

Therefore,

$A = \lceil \sqrt{3} \rceil + \lceil \sqrt{4}\rceil + \lceil \sqrt{5} \rceil + \ldots + \lceil \sqrt{83} \rceil \\ =2(2) + 3(5) + 4(7) + 5(9) + 6(11) + 7(13) + 8(15) + 9(17) + 10(2)\\ =540$

We then calculate the desired answer:

$\dfrac{1}{3} \lceil \sqrt{A} \rceil\\ =\dfrac{1}{3} \lceil \sqrt{540} \rceil\\ =\dfrac{1}{3} \lceil 23.24 \rceil\\ =\dfrac{1}{3} (24)\\ =\boxed{8}$