Let  ${ a\uparrow\uparrow b = {{{{{a^{a}}^a}^{\dots}}}^{a}}^{a}} $, where there are $ b $ a's in total. That is $ a\uparrow\uparrow b $ is given by the recurrence \[  a\uparrow\uparrow b = \begin{cases} a & b=1\\ a^{a\uparrow\uparrow (b-1)} & b\ge2\end{cases} \] What is the remainder of $  3\uparrow\uparrow( 3\uparrow\uparrow ( 3\uparrow\uparrow 3)) $ when divided by $ 60 $?
To find the remainder of \( 3 \uparrow\uparrow (3 \uparrow\uparrow (3 \uparrow\uparrow 3)) \) when divided by 60, we need to evaluate the expression step by step and use modular arithmetic properties.

1. **Evaluate \( 3 \uparrow\uparrow 3 \):**
   \[
   3 \uparrow\uparrow 3 = 3^{3^3} = 3^{27}
   \]

2. **Evaluate \( 3 \uparrow\uparrow (3 \uparrow\uparrow 3) \):**
   \[
   3 \uparrow\uparrow (3 \uparrow\uparrow 3) = 3 \uparrow\uparrow (3^{27})
   \]
   This is a very large number, so we need to use modular arithmetic to simplify it.

3. **Evaluate \( 3^{27} \mod 20 \):**
   \[
   3^3 = 27 \equiv 7 \mod 20
   \]
   \[
   3^6 = (3^3)^2 = 7^2 = 49 \equiv 9 \mod 20
   \]
   \[
   3^9 = 3^6 \cdot 3^3 = 9 \cdot 7 = 63 \equiv 3 \mod 20
   \]
   \[
   3^{27} = (3^9)^3 \equiv 3^3 = 27 \equiv 7 \mod 20
   \]

4. **Evaluate \( 3 \uparrow\uparrow 7 \mod 60 \):**
   \[
   3 \uparrow\uparrow 7 = 3^{3^{3^{3^{3^{3^3}}}}}
   \]
   We need to find \( 3^7 \mod 60 \):
   \[
   3^2 = 9
   \]
   \[
   3^3 = 27
   \]
   \[
   3^4 = 81 \equiv 21 \mod 60
   \]
   \[
   3^5 = 3 \cdot 21 = 63 \equiv 3 \mod 60
   \]
   \[
   3^6 = 3 \cdot 3 = 9 \mod 60
   \]
   \[
   3^7 = 3 \cdot 9 = 27 \mod 60
   \]

Thus, the remainder of \( 3 \uparrow\uparrow (3 \uparrow\uparrow (3 \uparrow\uparrow 3)) \) when divided by 60 is \( 27 \).

The final answer is \( \boxed{ 27 } \).