Given, $$Q = {{a{b^4}} \over {cd}}$$

$$a = (60 \pm 3)\,Pa \Rightarrow a = 60\,Pa,\,\Delta a = 3Pa$$

$$b = (20 \pm 0.1)\,m \Rightarrow b = 20m,\,\Delta b = 0.1\,m$$

$$c = (40 \pm 0.2)\,Ns{m^{ - 2}} \Rightarrow c = 40\,Ns{m^{ - 2}},\,\Delta c = 0.2\,Ns{m^{ - 2}}$$

$$d = (50 \pm 0.1)\,m \Rightarrow \,d = 50\,m,\,\Delta d = 0.1m$$

As, $$Q = {{a{b^4}} \over {cd}}$$

by taking ln on both sides,

$$\ln Q = \ln a + u\ln b - \ln c - \ln d$$

Now, by differentiating,

$${{dQ} \over Q} = {{da} \over a} + 4{{db} \over b} - {{dc} \over c} - {{dd} \over d}$$

So, maximum fractional error in Q is given by,

$${{\Delta Q} \over Q} = {{\Delta a} \over a} + 4{{\Delta b} \over b} + {{\Delta c} \over c} + {{\Delta d} \over d}$$

$$ \Rightarrow {{\Delta Q} \over Q} = {3 \over {60}} + 4\left( {{{0.1} \over {20}}} \right) + {{0.2} \over {40}} + {{0.1} \over {40}}$$

$$ = {1 \over {20}} + {1 \over {50}} + {1 \over {200}} + {1 \over {500}}$$

$$ \Rightarrow {{\Delta Q} \over Q} = {{50 + 20 + 5 + 2} \over {1000}} = {{77} \over {1000}}$$

Hence the % error in $$Q = {{\Delta Q} \over Q} \times 100\% $$

$$ = {{7700} \over {1000}}\% = {x \over {1000}}$$ (given)

So, $$x = 7700$$