How to size an in-line fuse for the input of the DC-DC power supply
The fuse must be chosen for both the steady state load and initial connection input connection with Meanwell’s spec. of  INRUSH CURRENT (Typ.) 45A/48VDC
The storage caps must be charged up and the I^2t must be measured to choose the appropriate size with ambient temp derating.  Fuses have a very wide tolerance to trip for inrush, and it is not easy to choose a slow-blow.  Too small makes it unreliable for frequent surge operations and too big makes it take a long time if it starts to draw twice it’s rated current.  But the capacitor ESR inside is what often limits that current surge and output ripple, so these are tradeoffs.
This is a 200W output supply that draws a surge peak of  ~ 48V x 45A = 2160 W but how many Joules of energy is stored to provide a high efficiency step down voltage is the key factor which is unknown.
The input current will be max. steady-state for the Vmin input which 36V * 6.7A = 242 W with Imax output of 200W indicating 80% efficient a Vmin.  Your efficiency ought be better at 48V , perhaps 85%.

a slo-blow fuse at twice the rating takes about 2 minutes to fuse at room temp.
an 8A time-delay fuse might be able to survive the exponentially decaying surge because it might take 100ms at 45 A to trip and the current at this time will be well below.
I suspect if the Joules of surge energy was known, and the fuse withstanding time delay energy was known, it could be a simple calculation to choose a safety margin.
In the end, I suspect an 8A to 10A fuse will blow before the battery or cable but not trip from the surge energy to start.
There may be clever ways to soft start if Meanwell agrees it won’t overheat any internal components