Suppose $t_1, t_2, ...t_5$ are in AP such that $\sum ^{18}_{l=0}{{t}}_{3l+1}=1197$ and ${{t}}_7+{{3}}t_{22}=174$. If $\sum ^9_{l=1}{{{t}}_l}^2=947b$, then the value of $b$ is

$= 19a + 3d\cdot\dfrac{18 \times 19}{2} $

$= 19a + 513d = 1197$

$\Rightarrow a + 27d = 63 \quad \cdots(1)$

Step 2: Use $t_7 + 3t_{22} = 174$
$t_7 = a + 6d,\quad t_{22} = a + 21d$

$(a + 6d) + 3(a + 21d) = 174$
$4a + 69d = 174 \quad \cdots(2)$

Step 3: Solve (1) and (2)
From (1): $a = 63 - 27d$

Substitute in (2):
$4(63 - 27d) + 69d = 174$
$252 - 108d + 69d = 174$
$-39d = -78$
$d = 2$

$a = 63 - 54 = 9$

Step 4: Find $\displaystyle\sum_{l=1}^{9} t_l^2$
$t_l = 9 + (l-1) \times 2 = 7 + 2l$

$\displaystyle\sum_{l=1}^{9} t_l^2 = \sum_{l=1}^{9}(7+2l)^2 = \sum_{l=1}^{9}(49 + 28l + 4l^2)$

$= 49(9) + 28\cdot\dfrac{9 \times 10}{2} + 4\cdot\dfrac{9 \times 10 \times 19}{6}$

$= 441 + 1260 + 1140$

$= 2841$

Step 5: Find $b$
$947b = 2841$

$b = \dfrac{2841}{947} = 3$

Answer: $b = \boxed{3}$